K223 / K224 / 2027
Energy overview

Lesson 2 of 4

Kinetic and gravitational energy

  • I can apply kinetic energy = 1/2 mv^2 and gravitational potential energy = mgh near Earth to new situations.Syllabus

    Syllabus K223 / K224, 5(b). Apply kinetic energy = 1/2 mv^2 and gravitational potential energy = mgh near Earth to new situations.

Make a guess. You are not marked.

A car speeds up from 10 m/s to 20 m/s. What happens to its kinetic energy?

  1. It doubles, because kinetic energy is directly proportional to the speed.
  2. It becomes four times as large, because it depends on speed squared.
  3. It stays the same, because the mass is unchanged.
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It becomes four times as large, because it depends on speed squared.

Kinetic energy is 1/2 m v2, so doubling the speed squares to a factor of four. The 20 m/s car has four times the kinetic energy.

In short

  • Kinetic energy: Ek = 1/2 m v2.
  • Gain in gravitational potential energy: Ep = mgh, using the vertical height.
  • Doubling the speed gives four times the kinetic energy.

Kinetic energy depends on mass and speed. A change in gravitational potential energy near Earth depends on mass and vertical height change.

Use mass in kg, speed in m/s and height in m. Gravitational field strength g is measured in N/kg. Review unit conversions before substituting.

Kinetic energy: Ek = 1/2 mv2m = mass in kg; v = speed in m/s. Ek is measured in J. Square the speed, not the whole expression.

Worked example

A moving 2.0 kg object

At a speed of 4.0 m/s:

Ek = 0.5 x 2.0 x 4.02 = 0.5 x 2.0 x 16 = 16 J.

At 8.0 m/s, the same mass has Ek = 0.5 x 2.0 x 64 = 64 J. Doubling the speed makes the kinetic energy four times as large.

Try it step by step

Work it out in steps: Speed in m/s, then Speed squared in m2/s2, then Kinetic energy in J.

  1. Problem 1. A car of mass 800 kg travels at 54 km/h. Find its kinetic energy. (1 km/h = 1000 m / 3600 s.)
  2. Problem 2. A van of mass 1200 kg travels at 72 km/h. Find its kinetic energy.
  3. Problem 3. A cyclist and bicycle have a total mass of 60 kg and travel at 36 km/h. Find the kinetic energy.
Show the answer

Problem 1.

Speed in m/s: v = 54 x 1000 / 3600 = 15 m/s

Speed squared in m2/s2: v2 = 15 x 15 = 225 m2/s2

Kinetic energy in J: Ek = 1/2 m v2 = 0.5 x 800 x 225 = 90000 J, which is 90 kJ (2 sf)

Problem 2.

Speed in m/s: v = 72 x 1000 / 3600 = 20 m/s

Speed squared in m2/s2: v2 = 20 x 20 = 400 m2/s2

Kinetic energy in J: Ek = 0.5 x 1200 x 400 = 240000 J, which is 240 kJ (2 sf)

Problem 3.

Speed in m/s: v = 36 x 1000 / 3600 = 10 m/s

Speed squared in m2/s2: v2 = 10 x 10 = 100 m2/s2

Kinetic energy in J: Ek = 0.5 x 60 x 100 = 3000 J (2 sf)

  • At fixed speed, doubling the mass doubles kinetic energy.
  • At fixed mass, doubling speed multiplies kinetic energy by four; halving speed makes it one quarter.
  • The direction of motion does not change this calculation. Use speed, and give energy in J without a spatial direction.

Use a vertical height change

Gravitational potential energy change = mghm = mass in kg; g = gravitational field strength in N/kg; h = vertical height change in m. The result is in J. Near Earth, take g as approximately constant over the height considered.

Raising a mass increases the gravitational store of the mass-Earth system; lowering it decreases the store. The numerical value Ep = mgh is measured relative to a chosen zero-height level. Changing that reference changes the assigned value, but not the energy change between the same two positions.

Worked example

Raise a load through 1.5 m

A total mass of 2.0 kg is raised vertically by 1.5 m. Use g = 10 N/kg.

Increase in Ep = 2.0 x 10 x 1.5 = 30 J.

Its weight is mg = 20 N. Lifting it to the same height along a longer ramp still increases the gravitational store by 30 J. Extra transfers due to friction would change the input required, not this gravitational change.

At fixed m and g, doubling the vertical rise doubles the gravitational store increase. At fixed rise and g, doubling the mass also doubles the increase. If the same load descends by 1.5 m, the store decreases by 30 J.

Rearrange for the quantity asked for

If the height change is unknown, h = change in Ep / (mg). For the 2.0 kg load, a 30 J increase at g = 10 N/kg gives h = 30 / (2.0 x 10) = 1.5 m.

If speed is unknown, first rearrange to v2 = 2Ek / m, then take the square root. Do not treat the value of v2 as the speed.

Try it step by step

Work it out in steps: Weight mg in N, then Height h = Ep / (mg) in m.

  1. Problem 1. A crane raises a 3.0 kg load and the gravitational potential energy increases by 450 J. Take g = 10 N/kg. Through what vertical height was the load raised?
  2. Problem 2. An 8.0 kg load gains 960 J of gravitational potential energy. Take g = 10 N/kg. Through what vertical height was it raised?
  3. Problem 3. A 5.0 kg load gains 275 J of gravitational potential energy. Take g = 10 N/kg. Through what vertical height was it raised?
Show the answer

Problem 1.

Weight mg in N: mg = 3.0 x 10 = 30 N

Height h = Ep / (mg) in m: h = 450 / 30 = 15 m (2 sf)

Problem 2.

Weight mg in N: mg = 8.0 x 10 = 80 N

Height h = Ep / (mg) in m: h = 960 / 80 = 12 m (2 sf)

Problem 3.

Weight mg in N: mg = 5.0 x 10 = 50 N

Height h = Ep / (mg) in m: h = 275 / 50 = 5.5 m (2 sf)

Height is not route length. Identify the starting and final vertical levels before using mgh. For kinetic energy, identify the speed at the instant being considered and square it.

Check your understanding

A 0.50 kg load is taken up a ramp of length 2.0 m. Its final position is 0.80 m vertically above its starting position. Use g = 10 N/kg. What is its increase in gravitational potential energy?

  1. 0.40 J
  2. 10 J
  3. 4.0 J
Show the answer

4.0 J

Use the vertical rise, 0.80 m: increase = mgh = 0.50 x 10 x 0.80 = 4.0 J. The route taken does not change this store increase.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A 0.50 kg ball is dropped from a height of 3.2 m. Air resistance is negligible and g = 10 N/kg. Describe the energy transfer as it falls, and calculate its speed just before it hits the ground. [3]

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